Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide. Ca) + H2O --> Ca(OH)2 In a particular experiment, a 5.00 g sample of CaO is reacted with excess water and 6.11 g of Ca(OH)2 is recovered. What is the percent yield in this experiment?

Respuesta :

Answer:

Percent yield = 92.5%

Explanation:

The question asks for the percent yield which can be defined as:

[tex]\frac{actual yield}{theoretical yield} .100[/tex]

Where the actual yield is how much product was obtained, in this case 6.11 g of Ca(OH)₂, and the theoretical yield is how much product could be obtained with the given reactants theoretically, that is if the reaction would work perfectly. So we need to calculate first the theoretical yield.

1. First lets write the chemical equation reaction correctly and check that it is balanced:

CaO + H₂O → Ca(OH)₂

2. Calculate the amount of product Ca(OH)₂ that can be obtained with the given reactants (theoretical yield), which are 5.00g of CaO and excess of water. So the amount of CaO will determined how much Ca(OH)₂ we can obtained.

For this we'll use the molar ratio between CaO and Ca(OH)₂ which we see it is 1:1. For every mol of CaO we'll obtain a mol of Ca(OH)₂. So lets convert the 5.00 g of CaO to moles:

 Molar Mass of CaO: 40.078 + 15.999 = 56.077 g/mol

 moles of CaO = 5.00 g / 56.077 g/mol = 0.08916 moles

As we said before from the molar ratio moles of Ca(OH)₂ = moles of CaO

So the moles of Ca(OH)₂ that can be obtained are 56.077 g/mol

We need to convert this value to grams:

 Molar Mass of Ca(OH)₂ = 40.078 + (15.999 + 1.008)*2 = 74.092 g/mol

Theoretical yield of Ca(OH)₂ = 0.08916 moles x 74.092 g/mol = 6.606 g

3. Calculate the percent yield:

[tex]\frac{actual yield}{theoretical yield} .100[/tex]

Percent yield = (6.11 g / 6.606g) x 100 = 92.5 %

The percentage yield of the experiment is 92.4%

The balanced equation for the reaction is given below:

CaO + H₂O —> Ca(OH)₂

Next, we shall determine the mass of CaO that reacted and the mass of Ca(OH)₂ produced from the balanced equation.

Molar mass of CaO = 40 + 16

= 56 g/mol

Mass of CaO from the balanced equation = 1 × 56 = 56 g

Molar mass of Ca(OH)₂ = 40 + 2(16 + 1)

= 40 + 2(17)

= 40 + 34

= 74 g/mol

Mass of Ca(OH)₂ from the balanced equation = 1 × 74 = 74 g

SUMMARY:

From the balanced equation above,

56 g of CaO reacted to produce 74 g of Ca(OH)₂.

Next, we shall determine the theoretical yield of Ca(OH)₂. This can be obtained as follow:

From the balanced equation above,

56 g of CaO reacted to produce 74 g of Ca(OH)₂.

Therefore,

5 g of CaO will react to produce = [tex]\frac{5 * 74}{56}[/tex] = 6.61 g of Ca(OH)₂.

Thus, the theoretical yield of Ca(OH)₂ is 6.61 g

Finally, we shall determine the percentage yield of Ca(OH)₂.

Actual yield of Ca(OH)₂ = 6.11 g

Theoretical yield of Ca(OH)₂ = 6.61 g

Percentage yield of Ca(OH)₂ =?

[tex]Percentage yield = \frac{Actual}{Theoretical} * 100[/tex]

Percentage yield of Ca(OH)₂ = [tex]\frac{6.11}{6.61} * 100[/tex]

Percentage yield of Ca(OH)₂ = 92.4%

Therefore, the percentage yield of the experiment is 92.4%

Learn more: https://brainly.com/question/14844156

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